ZOJ 3905 Cake

Cake

Time Limit: 4 Seconds Memory Limit: 65536 KB

Alice and Bob like eating cake very much. One day, Alice and Bob went to a bakery and bought many cakes.

Now we know that they have bought n cakes in the bakery. Both of them like delicious cakes, but they evaluate the cakes as different values. So they decided to divide those cakes by following method.

Alice and Bob do n / 2 steps, at each step, Alice choose 2 cakes, and Bob takes the cake that he evaluates it greater, and Alice take the rest cake.

Now Alice want to know the maximum sum of the value that she can get.

Input

The first line is an integer T which is the number of test cases.

For each test case, the first line is an integer n (1<=n<=800). Note that n is always an even integer.

In following n lines, each line contains two integers a[i] and b[i], where a[i] is the value of i^th cake that Alice evaluates, and b[i] is the value of i^th cake that Bob evaluates. (1<=a[i], b[i]<=1000000)

Note that a[1], a[2]..., a[n] are n distinct integers and b[1], b[2]..., b[n] are n distinct integers.

Output

For each test case, you need to output the maximum sum of the value that Alice can get in a line.

Sample Input

Sample Output

Author: HUA, Yiwei

解题：$动态规划，dp[i][j]表示有i个cake,选了j个给Bob，剩下的给Alice，由于预先按每块cake的b属性由高到低排序$

$所以，可以保证每次加入的Cake 如果是给Alice的一定可以成功的给Alice，因i里面有j个的cake的b属性比当前cake的b属性大$

 #include <bits/stdc++.h>

 using namespace std;

 const int maxn = ;

 typedef long long LL;

 struct Cake{

     int a,b;

     bool operator<(const Cake &rhs)const{

         return b > rhs.b;

     }

 }cake[maxn];

 LL dp[][maxn];

 int main(){

     int kase,n;

     scanf("%d",&kase);

     while(kase--){

         scanf("%d",&n);

         for(int i = ; i <= n; ++i)

             scanf("%d%d",&cake[i].a,&cake[i].b);

         sort(cake + ,cake + n + );

         memset(dp,,sizeof dp);

         int cur = ;

         for(int i = ; i <= n; ++i){

             for(int j = ((i-)>>); j >= ; --j){

                 dp[cur^][i-j] = max(dp[cur^][i-j],dp[cur][i--j]);

                 dp[cur^][i-j-] = max(dp[cur^][i-j-],dp[cur][i--j] + cake[i].a);

             }

             memset(dp[cur],,sizeof dp[cur]);

             cur ^= ;

         }

         printf("%lld\n",dp[cur][n>>]);

     }

     return ;

 }

巴特西

ZOJ 3905 Cake

Input

Output

Sample Input

Sample Output

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